Exercise 10.4 · Q10: Area of the parallelogram on $\hat i-\hat j+3\hat k$ and $2\hat i-7\hat j+\hat k$

Basics you need

Solution

1. The area is $|\vec a\times\vec b|$

   With $\vec a$ as the base, the height of the parallelogram is $|\vec b|\sin\theta$⁠, so

   $$\text{area} = |\vec a||\vec b|\sin\theta = |\vec a\times\vec b|.$$
   The height is the distance from the tip of $\vec b$ to the line of $\vec a$⁠, the side of a right triangle opposite the angle $\theta$ with hypotenuse $|\vec b|$⁠. By Definition 3, $|\vec a||\vec b|\sin\theta$ is exactly the length of $\vec a\times\vec b$⁠.
   Area of a parallelogram $= |\vec a\times\vec b|$ (Section 10.6.3, Observation 9). Lesson: the cross product

2. The vector product

   $$\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} \\ &= \hat i(-1 + 21) - \hat j(1 - 6) \\ &\quad + \hat k(-7 + 2) \\ &= 20\hat i + 5\hat j - 5\hat k. \end{aligned}$$

   Check: $(20\hat i + 5\hat j - 5\hat k)\cdot\vec a = 20 - 5 - 15 = 0$ and $(20\hat i + 5\hat j - 5\hat k)\cdot\vec b = 40 - 35 - 5 = 0$⁠.

   The minors are $(-1)(1) - (3)(-7) = -1 + 21 = 20$⁠, $(1)(1) - (3)(2) = 1 - 6 = -5$ and $(1)(-7) - (-1)(2) = -7 + 2 = -5$⁠; the $\hat j$ term takes a minus sign, giving $+5\hat j$⁠. Both scalar products are 0, as they must be for a vector perpendicular to both sides.
   Expansion along the top row, signs $+, -, +$⁠. Basics: a determinant measures a signed area

3. Its length

   $$\begin{aligned} |\vec a\times\vec b| &= \sqrt{20^2 + 5^2 + (-5)^2} \\ &= \sqrt{400 + 25 + 25} = \sqrt{450}. \end{aligned}$$

   Since $450 = 225 \times 2 = 15^2 \times 2$⁠, $\sqrt{450} = 15\sqrt2$⁠. The area is $15\sqrt2$ sq. units.

   Squaring removes the signs. Taking out the largest square factor, 225, gives the simplest form, the one in the answer key.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

4. Check with the scalar product, and the picture

   $|\vec a|^2 = 1 + 1 + 9 = 11$⁠, $|\vec b|^2 = 4 + 49 + 1 = 54$ and $\vec a\cdot\vec b = 2 + 7 + 3 = 12$⁠, so

   $$\begin{aligned} |\vec a\times\vec b|^2 &= |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 \\ &= 11 \times 54 - 12^2 \\ &= 594 - 144 = 450. \end{aligned}$$
   This route uses only the scalar product, so it checks the determinant independently. In the picture the longer side $\vec b$ is the base; the area does not depend on which side is the base, because $|\vec b\times\vec a| = |\vec a\times\vec b|$⁠.
   $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2$⁠, from $\sin^2\theta + \cos^2\theta = 1$ (Sections 10.6.1 and 10.6.3). Lesson: Examples 22–25

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